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2 years ago

HELP PLEASE I need to finish this asap

Physics

2 answers:

ELEN [110]2 years ago

6 0

Answer:

I'm not 100% sure, but I think the answer would be the first one because there's a force pushing the object in every direction, so they would cancel eachother out and make the object stay in the same place.

Explanation:

pls vote brainliest

kiruha [24]2 years ago

4 0

Third one bro I know this already

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A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal ran

marshall27 [118]

Answer

a) For the rock

$\dfrac{v_t^2sin 2\theta}{g} = \dfrac{v_t^2sin^2\theta}{2g}$

$2sin\thetacos\theta = \dfrac{sin^2\theta}{2}$

$2cos\theta = \dfrac{sin\theta}{2}$

$tan\theta = 4$

$\theta = tan^{-1} 4$

$\theta = 76^0$

b) $\theta = 45^0$ for maximum range

$\dfrac{d_{max}}{d}=\dfrac{(v_tcos 45^0)(2v_tsin 45^0)g}{(v_tcos 76^0)(2v_tsin 76^0)g}$

$\dfrac{d_{max}}{d}=\dfrac{0.707\times 0.707)}{0.97\times 0.242}$

$\dfrac{d_{max}}{d}=2.129$

$d_{max}=2.129 d$

c) The value of θ is the same on every planet as g divides out.

5 0

3 years ago

[O.04H]The table below shows the use of some energy production methods over time.

mr Goodwill [35]

I think The coastal areas were highly polluted

8 0

3 years ago

Read 2 more answers

a ball is thrown vertically upward with an initial speed of 40 m/s. how high is the ball above the ground when it stops

NISA [10]

Answer:

80m, assuming g=10m/s^2

Explanation:

40m/s will be reduced to 0m/s in 4 seconds. 4 seconds x 40m/s would be 160m up, but you will only get half of that because you decelerate linearly to 0m/s. This leaves you with 4 x 20 = 80m.

5 0

3 years ago

Read 2 more answers

An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of

Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, $E=390\ keV=390\times 10^3\ eV$

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2meE}}$

$\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}$

$\lambda=1.96\times 10^{-12}\ m$

(b) According to Bragg's law,

$n\lambda=2d\ sin\theta$

n = 1

For nickel, $d=0.092\times 10^{-9}\ m$

$\theta=sin^{-1}(\dfrac{\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})$

$\theta=0.010^{\circ}$

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0

3 years ago

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

4 0

3 years ago